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Triss [41]
2 years ago
5

50 trillion (5.00 x 1013) Angstrom is equivalent to 10900 cubit. If 108 Angstrom = 1 cm (exactly), how many m are there in 1.00

cubit?
Chemistry
1 answer:
Yuri [45]2 years ago
7 0

The number of meter (m) in 1 cubit, given the data is 4.25×10⁵ m

<h3>Data obtained from the question</h3>

From the question given above, the following data were obtained:

  • 50 trillion (5.00 x 1013) Angstrom = 10900 cubit
  • 108 Angstrom = 1 cm
  • How many meter (m) = 1 cubit?

<h3>How to convert 5×10¹³ Angstrom to cm</h3>

108 Angstrom = 1 cm

Therefore,

5×10¹³ Angstrom = 5×10¹³ / 108

5×10¹³ Angstrom = 4.63×10¹¹ cm

<h3>How to convert 1 cubit to cm</h3>

10900 cubit = 5×10¹³ Angstrom = 4.63×10¹¹ cm

10900 cubit = 4.63×10¹¹ cm

Therefore,

1 cubit = 4.63×10¹¹ / 10900

1 cubit = 4.25×10⁷ cm

<h3>How to convert 4.25×10⁷ cm to m</h3>

100 cm = 1 m

Therefore,

4.25×10⁷ cm = 4.25×10⁷/ 100

4.25×10⁷ cm = 4.25×10⁵ m

Thus,

1 cubit = 4.25×10⁷ cm = 4.25×10⁵ m

1 cubit = 4.25×10⁵ m

Learn more about conversion:

brainly.com/question/2139943

#SPJ1

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FrozenT [24]

Answer:

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(b) 11,625 mg/m3

Explanation:

The concentration of CO in mg/m3 can be calculated as

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For standard conditions (1 atm and 25°C), P/RT is 0.0409.

Concentration of 1.5% percent by volume of CO is equivalent to 1.5*10,000 ppm= 15,000 ppm CO.

The molecular weigth of CO is 28 g/mol.

(1) For 25°C and 1 atm conditions

C=(P/RT)*MW*C_{ppm}\\\\C=0.0409*28*15,000=17,178

(b) For 200°C and 1.1 atm,

P/RT=0.0409*(P/P_{std})*(T_{std}/T)\\P/RT=0.0409*(1.1atm/1atm)*(273+15K/273+200K)=0.0277

Then the concentration in mg/m3 is

C=(P/RT)*MW*C_{ppm}\\\\C=0.0277*28*15,000=11,625

5 0
3 years ago
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Balance the following reaction in acidic solution:. . Ag(s) + NO3-(aq) -&gt; Ag+(aq) + NO(g)
nekit [7.7K]
The two half-reactions are...
Ag→Ag+
and...
NO3→NO
Let's start by balancing the first half-reaction...
Ag→Ag+
The amounts are already balanced; 1:1. The oxygens are balanced. So all that's left is to balance the charge...
Ag→Ag++e−
Now let's do the other equation... Amounts of nitrogen are balanced, so we first need to balance the oxygens...
NO3→NO
4H++NO3→NO+2H2O
Next, we need to balance charge...
4e−+4H++NO3→NO+2H2O
Now let's go ahead and rewrite each half-reaction after being balanced by themselves...
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3 0
3 years ago
A 35.0-ml sample of 0.20 m lioh is titrated with 0.25 m hcl. What is the ph of the solution after 23.0 ml of hcl have been added
kiruha [24]
<h3><u>Answer;</u></h3>

pH = 12.33

<h3><u>Explanation;</u></h3>

The equation of reaction is :

LiOH(aq) + HCl(aq) --> LiCl(aq) + H2O(l)

Reactants left after the titrant is added;

Total Moles LiOH;

= 0.035L LiOH × (0.2moles/L)

= 0.007moles of LiOH

Moles of HCl;

= 0.023L HCl × (0.25moles/L)

= 0.00575moles HCl is the limiting reagent

Reacting amount of moles of LiOH;

= 0.0575 moles HCl *(1mole LiOH/1moles HCl)

=0.00575 moles LiOH (reacted)

Moles of LiOH left;

= 0.007moles total - 0.00575moles that react

= .00125 moles of LiOH (left)

LiOH is a strong base, which means that it ionizes completely.  

0.00125moles LiOH *(moles/0.058L) = 0.02155M of LiOH

LiOH(aq) --> Li+(aq) + OH-(aq)

[LiOH] = [OH-] = 0.02155 M

pOH = -log[OH-]

pOH = -log(0.02155)

pOH= 1.67

pH = 14 - pOH

pH = 14 - 1.67

pH = 12.33

7 0
3 years ago
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